If α, β are roots of the equation x ^ 2 + 5( 2 )x + 10 = 0, α > β...

Given, ${x}^{2} + 5\sqrt{2}x + 10 = 0$

and ${P}_{n} = {\alpha}^{n} - {\beta}^{n}$

Now $\dfrac{{P}_{17}{P}_{20} + 5\sqrt{2}{P}_{17}{P}_{19}}{{P}_{18}{P}_{19} + 5\sqrt{2}{P}_{18}^{2}} = \dfrac{{P}_{17}\left({P}_{20} + 5\sqrt{2}{P}_{19}\right)}{{P}_{18}\left({P}_{19} + 5\sqrt{2}{P}_{18}\right)}$

$= \dfrac{{P}_{17}\left({\alpha}^{20} - {\beta}^{20} + 5\sqrt{2}\left({\alpha}^{19} - {\beta}^{19}\right)\right)}{{P}_{18}\left({\alpha}^{19} - {\beta}^{19} + 5\sqrt{2}\left({\alpha}^{18} - {\beta}^{18}\right)\right)}$

$= \dfrac{{P}_{17}\left({\alpha}^{19}\left(\alpha + 5\sqrt{2}\right) - {\beta}^{19}\left(\beta + 5\sqrt{2}\right)\right)}{{P}_{18}\left({\alpha}^{18}\left(\alpha + 5\sqrt{2}\right) - {\beta}^{18}\left(\beta + 5\sqrt{2}\right)\right)}$

Since $\alpha + 5\sqrt{2} = - 10 / \alpha . . . \left(1\right)$

$\text{ and } \beta + 5\sqrt{2} = - 10 / \beta . . . \left(2\right)$

Now put these values in above expression

$\dfrac{{P}_{17}\left({\alpha}^{19}\left(\alpha + 5\sqrt{2}\right) - {\beta}^{19}\left(\beta + 5\sqrt{2}\right)\right)}{{P}_{18}\left({\alpha}^{18}\left(\alpha + 5\sqrt{2}\right) - {\beta}^{18}\left(\beta + 5\sqrt{2}\right)\right)} = \dfrac{- 10{P}_{17}{P}_{18}}{- 10{P}_{18}{P}_{17}} = 1$